Question

How many grams `H_(2)SO_(4)` can be obtained from `1320gm PbS` as per reaction sequence ?.
`2PbS+O_(2)rarr2PbO+2SO_(2)`
`3SO_(2)+2HNO_(3)+2H_(2)Orarr3H_(2)SO_(4)+2NO`
`["At mass" Pb = 208. S = 32]`.

Answer 1

Moles of PbS=`(1320)/(240)=5.5` mol
Moles of `SO_(2)=5.5` mol =moles of `H_(2)SO_(4)`
Mass of `H_(2)SO_(4)=5.5xx98=539` gm
[When amount of only one reactant is given generaly other is assumed in excess.